Question

The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

• A. $8\sqrt{3}$
• B. 12
• C. $5\sqrt{5}$
• D. $10\sqrt{2}$

Answer 1

Answer: (D)

$10\sqrt{2}$

Answer 2

From the diagram, it is evident that AB serves as both the height of the equilateral triangle and the slant height of the pyramid.

So, $AB=\frac{\sqrt{3}}{2}×side=\frac{\sqrt{3}}{2}×20=10$

And $AO=\frac{1}{2}×side=\frac{1}{2}×20=10$
Applying Pythagoras theorem in triangle AOB

$OB^2=AB^2−OA^2$
$=(10\sqrt{3})^2−10^2$
$=200$

Hence, the height of the pyramid $(OB) =10\sqrt{2}$